The given expression
5−(3a2−2a)(6−3a2+2a) can be factorized as follows:
5−(3a2−2a)(6−3a2+2a)=5−(3a2−2a)[6−(3a2−2a)]=5−y(6−y)(Assuming3a2−2a=y)=5−6y+y2=y2−6y+5=y2−5y−y+5=y(y−5)−1(y−5)=(y−5)(y−1)=(3a2−2a−5)(3a2−2a−1)(∵3a2−2a=y)=(3a2+3a−5a−5)(3a2−3a+a−1)=[3a(a+1)−5(a+1)][3a(a−1)+1(a−1)]=(3a−5)(a−1)(a+1)(3a+1)
Hence, the factorization of 5−(3a2−2a)(6−3a2+2a) is (3a−5)(a−1)(a+1)(3a+1).